Transcript of "Orbital Mechanics: 6. Complications To Impulsive Maneuvers"
1.
6 Complications to Impulsive Maneuvers Having considered one-, two-, and three-impulse A Voptimal transfers betweenorbits in Chapter 5, we now turn our attention to the complications of the real world.Are there cases in which four or more impulses would offer significant savings?Do we always want to use the AV optimal transfer? How bad is our assumptionof impulsive AV application, and how does it affect the results? These are thequestions that will be investigated in this chapter. Specifically, the following topics will be considered: 1) N-impulse maneuver,2) fixed-impulse transfers, 3) finite-duration burns, and 4) very low thrust transfers.6.1 N-Impulse Maneuvers In previous chapters, we have seen cases in which a two-impulse maneuvercan accomplish the same transfer as a single-impulse but at a considerable AVsaving (e.g., argument of perigee change). Similarly, cases were studied in whicha three-impulse maneuver was best (e.g., bi-elliptic transfers). The obvious nextquestion is: What about four or more impulses? Edelbaum ~ has answered this question exactly in his paper "How Many Im-pulses?" The answer, he finds, is that only in rare situations are more than threeimpulses required to obtain the minimum A V transfer.6.2 Fixed-Impulse TransfersIntroduction To date, nearly all of the satellites that have been placed in low Earth orbit by theSpace Shuttle have achieved their final mission orbit using solid-propellant rocketmotors. Whether they are inertial-upper-stage (1US) or payload-assist-module(PAM) engines, their operation is basically the same. Each solid rocket motorcan be viewed simply as a container of solid propellant with a nozzle. An igniterbegins the burning of the solid propellant, which continues until all propellant hasbeen consumed. Whereas it is technically feasible to quench the burning of thesolid propellant, the typical solid motors in use today are neither stoppable norrestartable. Since most orbital transfers call for two burns, the typical upper-stage vehiclemust consist of two solid motors as depicted in Fig. 6.1. Stage I provides the firstburn (AV1) and stage II provides the second burn (AV2). For a given payload weight and fixed (off-the-shelf) solid rocket motors, thea V available from the two motors are fixed by the rocket equation as derived inChapter 1: Wi -- = exp( A V/go Isp) (6.1) w: 117
2.
118 ORBITAL MECHANICS Stage II SOLID ROCKET MOTORS Stage I Fig. 6.1 Schematic diagram of a payload atop two solid rocket motors.where go = the acceleration of gravity at the Earths surface Isp = engine specific impulse Wi = weight of rocket before burn (initial) Wf = weight of rocket after burn (final) Since the solid motors cannot be extinguished or restarted, the A I/1 and A V2supplied by the motors must exactly match the orbital-transfer A V requirements.The problem becomes one of finding a transfer between two given orbits thatexactly matches the available A V1 and A I12. This problem is often referred to asthe "velocity-matching technique" or "fixed-impulse transfers" (referring to thefact that the impulse available from each engine is fixed). This is clearly a complication to the process of selecting an orbital transfer.In earlier chapters, the orbital transfer was sought that would minimize total A Vexpenditure. The inherent assumption was that the A V could be metered out inany size increments by the rocket engine. Consider the plight of the designer who must assemble a pair of solid motors toform an upper-stage vehicle. In many cases, a range of different payload weightsis to be carried, even if it is just to allow for a growth version of a single satellite.In this event, the designer must size the motors for the heaviest payload expected.Having done this, he must find a method to fly the smaller payloads with the samemotors. Even the designer who has only a single satellite weight to contend withmight be required to select his solid motors from the existing, off-the-shelf array.It is unlikely that he will find motors that will exactly match the A V of the optimaltransfer. In any event, the designers problem is the same: how to utilize solidmotor stages where either one or both of the stages have excess energy above theoptimal-transfer A V requirements. The problem is always excess energy because,if the sum of the A V provided by the solid motors is less than that of the optimaltransfer, no solution is possible. Several methods have been identified to accommodate this excess energy: 1) Offloading propellant. 2) Adding ballast. 3) Trajectory modification.
3.
COMPLICATIONS TO IMPULSIVE MANEUVERS 119 The first method, offloading propellants, involves the custom removal of solidpropellant from the engine. Propellant is usually removed in the order in which theburn would normally proceed so as to impact the burn characteristics of the motoras little as possible. The disadvantages of this method are that custom propellantoffloading costs money and, for large offioads, may require requalifying the motorfor flight. An altemate method consists of adding ballast weights to the appropriate stagesto make the A V from the motors match the optimal-transfer requirements. Alimitation of this approach is that, once enough weight has been added to matchthe second-stage A V to its transfer requirement, the first stage may no longer haveenough AV. A second drawback is that bolting heavy weights to the payload ormotors may invalidate previously performed structural and vibrations tests. The third method, trajectory modification, is the one we will examine in detail.In trajectory modification, rather than alter the stage AV to match the optimal-transfer A V, we alter the orbit transfer to match the stage A V. The orbit transfer isno longer a A V optimal transfer. Physically speaking, the excess energy is beingused up via combinations of extra plane change, altitude change, and flight-path-angle change. Usually, there is not a unique solution but rather a range of solutionsavailable to choose from. The following sections will illustrate the trajectorymodification method with an example. For more details on this subject, Refs. 2and 3 are recommended.Fixed-Impulse Transfer Example Problem statement. Consider a constellation of satellites that operate incircular orbits with a 12-h period. As shown in Fig. 6.2, the satellite is deliveredto low Earth orbit (altitude hi = 278 km) by the Space Shuttle. From there itis carried to its destination orbit (h2 = 20187 kin, inclination i = 55 deg) by anupper-stage vehicle consisting of a pair of solid rocket motors. The arrangement and characteristics of these motors are shown in Fig. 6.3.Using the rocket equation to evaluate the A V that are available from each stage AV2 FINAL ORBIT h2 = 20186 km i 2 = 55 D E G SHUTTLE ORBIT h I = 278 km i I = VARIOUSFig. 6.2 Transfer from Shuttle orbit to final orbit is accomplished by an upper stage(consisting of two motors) attached to the payload.
4.
120 ORBITAL MECHANICS WEIGHTS (kg) 907 Rocket equation on second burn: STRUCTURE 91 W....... = 907 + 91 + 898 = exp(AV2/golsp) W~i.,,~ 9()7 + 91 PROPELLANT 898 AV2 = 1888 m/s / Rocket equation on first burn: STRUCTURE ] 181 W..it., I _ 91)7 + 91 + 898 + 181 + 2174 Wli,,~i 907 + 91 + 898 + 181 PROPELLANT l 2174 exp(AVd go lsp) AV, - 211)7 m/s / Fig. 6.3 Upper-stage vehicle (/SP for both motors = 300 s; go = 9.8066 m/s2).yields the results AV1 = 2107 m/s and AV2 = 1888 m/s. Our objective is to findorbit transfers that utilize exactly these amounts of A V. Hohmann transfers. Hohmann-type transfers can still be used if the amountof plane change at both burns is custom-tailored to match the A V capabilities of thesolid motors. The transfer orbits A V1 and A V2 are derived in the diagram shownin Fig. 6.4 as a function of the amount of plane change (C~l and ~2) performed onthe first and second bums, respectively. In these equations, r] = radius of Shuttle orbit r2 = radius of final orbit Vc1 = circular velocity of Shuttle orbit /" N t SECOND BURN (EOMETRY V ATR ~ Vc2 (6.2) AV22 = V2c2 + VATR2 _2Vc2VATRCOSa2 FIRST BURN GEOMETRY ~Avl Vcl (6.3) AVI2 = V2cl + VpTR2 _2VclVpTRCOS % Fig. 6.4 Hohmann transfer with plane change split.
5.
COMPLICATIONS TO IMPULSIVE MANEUVERS 121 Vc2 = circular velocity of final orbit aTR = semimajor axis of transfer orbit VprrR = transfer-orbit perigee velocity VATR= transfer-orbit apogee velocityFor these solid motors, we know AVI = 2107 m/s AV2 = 1888 m/sSolving for the above quantities, we find rl = hi -~ re = 6656 km r2 = h 2 + re = 26565 kin re = Earth radius Vd = ~r~/ra = 7739 m/s gc2 = ~r~/r2 = 3874 m/s (rl + r2) aZR-- - - -- 1 6 6 1 1 km 2 VPTR = 9787 m/s VLR---- ::ff V A T R ----- 2 4 5 2 m/sSolving the equations in Fig. 6.4 for o/t, 0/2, we find 0/1 = 3.25 deg o~2 = 23.25 deg Thus, if these specific amounts of plane change are performed as part of aHohmann ascent, all propellant will he consumed in both motors. What the equa-tions do not specify is the direction in which the plane change is performed. This isfree for us to choose. If the two plane changes are performed in the same direction,then the plane change angle 0 between the Shuttle orbit and the final orbit mustbe as follows: Plane change in same direction: 0 a = O/1 Jr- 0/2 = 3.25 + 23.25 = 26.5 degHowever, if the two plane changes are performed in opposing directions, then thetotal plane change angle between the two orbits is only as follows: Plane changes in opposing directions: Ob = 10/1 - - 0/21 = 23.25 - 3.25 = 20 deg
6.
122 ORBITAL MECHANICSIf all the plane change is performed as inclination change, then the Shuttle orbitcould be placed at any one of four different inclinations ia: il : i2 -4- 0a = 28.5, 81.5 : i2 -4- Ob : 35, 75 In general, however, plane change consists of both inclination change and rightascension of ascending node change (Ag2) according to the relation cos 0 : cos il cos i2 + sin il sin i2 COS A ~ (6.4)Substituting 0 = 0a into this equation, varying 28.5 < il < 81.5, and solving forA~~ a a s a function of il will produce all Shuttle orbit orientations (Af2, i) that areexactly 0, away from the final orbit in terms of total plane change. This has beendone, and the results are plotted in Fig. 6.5. Thus, from each point on the 0a curve,a Hohmann transfer can be accomplished to the final orbit, and the individualplane changes on each burn (oq = 3.25 deg, de = 23.25 deg) will cause eachsolid motor to be exhausted exactly. Similarly, substituting 0 -- Ob in the relation(6.4) yields the second curve in Fig. 6.5. Again, any Shuttle orbit orientation onthe Ob curve, specified as il, Af2 (where Af2 = 0 when the orbit has the sameright ascension of ascending node as the final orbit) is a viable orbit in terms ofexactly matching the A V using a Hohmann transfer. In the case of Ob, however, theplane changes are in opposing directions. That is, one burn raises the inclination,whereas the other decreases it. Figure 6.5 then contains all possible locations for 85 l,ot:i o f IarkD~g O r b i t w h i c h m a t c h AV I a n d AV2 a v a i l a b l v . 75 65 Plane changes i n i same d i r e c t i o l ~ , z 55 - ~ Plane changes t n opposing d i r e c t i o n . J mission 25 -35 -25 -15 -5 5 15 25 35 CHANGE IN RIGHT ASCENSION OF ASCENDING NODE {DEG) Fig. 6.5 Hohmann flight time transfers.
7.
COMPLICATIONS TO IMPULSIVE MANEUVERS 123 I h2~,. Fig. 6.6 Non-Hohmann transfers.the Shuttle orbit if the trajectory modification method is to be used to match AVand a Hohmann transfer is desired. Non-Hohmann transfers. But there is nothing sacred about Hohmann trans-fers. Consider what happens as the transfer-orbit apogee altitude is raised be-yond the final orbit altitude. As shown in Fig. 6.6, as the transfer orbit apogeealtitude is increased beyond the Hohmann-transfcr value, two things happen:1) 0/1 decreases because VpTR increases; and 2) 0/2 decreases because an increas-ing flight-path-angle change T2 must be performed. To reach the higher apogeealtitude, the perigee velocity in the transfer orbit (VpTR) must increase. From Eq.(6.3), for a fixed AV1 and Vd as VpTR increases, 0/1 must decrease. Similarly, 0/2decreases since an increasing flight-path-angle change T2 must also be performed.The apogee altitude of the transfer orbit can be increased to the point at whicheither 0/1 or 0/2 reaches zero. Thus, as transfer-orbit apogee altitude is increased,Oa decreases, and Ob approaches 0a. In the limiting case, Ob will equal 0a. (Notethat 0a and 0 b are no longer the simple sum and difference of 0/1 and 0/2 as aresult of different axes of rotation. For a more detailed explanation, see Refs. 2and 3). Figure 6.7 shows our example for a non-Hohmann transfer orbit with apogeealtitude very near its upper limit. These have been labeled "max/min flight timetransfers." The minimum flight time is associated with the ascending crossing ofthe final orbit, and the maximum flight time is associated with the descendingcrossing. Again, these curves show, for a specific value of transfer-orbit apogeealtitude, all Shuttle orbits from which the final orbit can be reached using A Vl =2107 m/s and AV2 = 1888 rigs. In Fig. 6.5, the transfer-orbit apogee altitude isthat of the Hohmann transfer. In Fig. 6.7, the transfer-orbit apogee altitude is very
8.
124 ORBITAL MECHANICS 85. Loci of ParkingOrbits which /match AV1 and AV2 available. 75. 65. N n Plane changes in same direction. 55. IL Plane changes in opposing direction. < z 35 25. -35. -25. -15. -5, 5. 15. 25, 3m~l~sion~ orbit CHANGE IN RIGHTASCEN$1ONOFASCEND1NG NODE (DEG) t r a n s f e ~ orbit parking orb Fig. 6.7 Min/max flight time transfers.near its maximum allowable value, as evidenced by the closeness of the 0a and 0bcurves. Composite trajectory modification results. As the transfer-orbit apogeealtitude is raised from the Hohmann value (a lower limit) to the maximum value,the Oa and 0b curves traverse the area between Figs. 6.5 and 6.7. The compositeresult for all possible transfer-orbit apogee altitudes is shown in Fig. 6.8. Froma Shuttle orbit located anywhere within the shaded area of Fig. 6.8, at least onetransfer exists that satisfies the fixed values of A V1 and A V2 for the final example.Only those points that are on the curves of Fig. 6.5 will have a Hohmann-transfercapability. All others involve non-Hohmann-type transfers. Note that the outerboundary of the region consists of Hohmann transfers, whereas the inner boundary(if any) consists of max/min flight-time solutions. For Shuttle orbits outside theregion of Fig. 6.8, there is insufficient AV in the solid motors. For Shuttle orbitsinside the inner bound of Fig. 6.8, there is too much AV in the solid motors foreven the trajectory modification method to handle. At a point within the shadedregion, usage of the trajectory modification method may yield several choices oftransfer orbit that satisfy the A V constraints.Launch Window It would seem that by supplying excess energy above and beyond the AVoptimal transfer, we should be receiving some benefit in return. In fact, we doreceive a benefit called a launch window, that is, a duration of time within whicha ground launch to the Shuttle orbit is possible.
9.
COMPLICATIONS TO IMPULSIVE MANEUVERS 125 85.Locl of Parking Orbits(shaded) I;o match ~V I - ) .-J.-.L..J...I-. I i___i_..]_..dj...and Z~/2 s v a i l a b l e . " I l l * i i i i 75. . . . . . . . i--i--- :I :: ess- i--- i i 35. I . . . . . . _ i ] I I I I l l 2s. 1 ~ 1 i -35. -25. -15. -5. 5. 15. 25. 3S. CHANGE IN RIGHT ASCENSION OF ASCENDING NODE (DEG)Fig. 6.8 Regionofallowableorbits. Missionorbit:h=20,186km, i=55deg,~=Odeg; parking orbit: h = 278 km, A V 1 = 2107 m/s, A V 2 = 1888 m/s. Because the Earth rotates eastward at approximately 15 deg/h (360 deg/23 h,56 min, 4.09 s), if a launch takes place t hours late, then the right ascension of theascending node of the orbit will be = f2o + 15twhere f2 = actual value of right ascension of ascending node attained f20 = target value of right ascension of ascending node The effect of an early or late launch on orbit-transfer geometry is shown inFig. 6.9. The acceptable amount of time by which a launch may be early or lateis called the launch window. For our example, if the Shuttle orbit is located ati = 55 deg, the launch window is shown on Fig. 6.8. In terms of Af2 and time(where t = 0 is the time the final orbit plane passes over the launch site), thelaunch window extends from Af2 (deg) - 3 2 to - 1 7 17 to 32 t(h) - 2 . 1 3 to - 1 . 1 3 1.13 to 2.13Note that, because of excess energy constraints, the region around the conodallaunch time (t = 0) is not part of the launch window. (Conodal launch time iswhen shuttle and final orbit f2 are the same, A~2 = 0.) Similarly, for a Shuttle orbit at il----35 deg, the launch window is t =0 + 1.73 h. In this case, the conodal launch time is within the launch window.
10.
126 ORBITAL MECHANICS ION ORBIT ! !" ...PARKING ORBIT FOR EARLY LAUNCH ..: ISFER ORBIT OARKING ORBIT FOR LATE LAUNCH DIRECTION OF EARTH ROTATION Fig. 6.9 Effect of launch time on orbit geometry.For a Shuttle orbit at il = 28.5 deg, the launch window is simply an instant att = 0 when the final orbit plane is directly above the launch site. In all cases, as long as the Shuttle launch takes place within the allowablelaunch window, a transfer trajectory can be found that will exactly match the AVavailable from the solid motors.6,3 Finite-Duration Burns: Gravity LossesIntroduction All our work to this point has assumed "impulsive maneuvers"; that is, therequired A Vare applied instantaneously. To do this would require a propulsionsource of infinite thrust. In reality, the ratio of thrust to weight (T/W) is generallyin the range 10 -5 < T~ W < 10. This finite thrust to weigh ratio complicates ourprevious study of impulsive maneuvers by introducing a A V penalty, the so-calledgravity loss.Gravity Losses Consider a mass m traveling at velocity v. At some point in time, it ejects amass dm backward at a relative velocity u.
11.
COMPLICATIONS TO IMPULSIVE MANEUVERS 127 Before bum: Q 1) ) After burn:It should be noted that the specific impulse of a rocket engine is a measure of theexhaust velocity of the particles according to the relation u = g01spThe momentum balance equation for this exchange was derived in Eqs. (1.5-1.8)as dv dm m~ =-u--d-~-+Fwhere F is the sum of all extemal forces acting on the masses. The external forces acting on a launch vehicle are shown in Fig. 6.10. In thissimple model, the thrust T and the drag D act along the axis of the vehicle whilethe weight W of the vehicle acts in the downward direction. The flight-path angleof the vehicle has been labeled T. Making the appropriate substitutions in themomentum equation yields dv dm m-- = -u-- - D - mg sin T dt dt Multiplying by dt/m, we obtain dm D dv = - u - - - --dt- gsinTdt m m T D / W = Mg Fig. 6.10 External forces acting on launh vehicle.
12.
128 ORBITAL MECHANICSIntegrating from tl to t2 (duration of the bum) yields* dv = AV = --u -- --dt - g sinTdt l mor A V = g0Isp ~mfmi _ dt - g sin T d t (6.5)where mi = mass before bum (initial) m y = mass after bum (final) The result is the familiar rocket equation [Eq. (1.10)], plus two complicatingterms. In both cases, these terms detract from the rocket equation and reduce theavailable characteristic A V from the rocket. The first term is related to the dragforce D and is not of interest here. In subsequent discussion, we shall assumeD=0. The second term, the gravity-loss term, is the one we are interested in. Wecannot easily integrate this term since both g (the gravitational acceleration at therockets location) and T (the flight-path angle) are unspecified functions of thetime (which depend on the trajectory flown). We can, however, analyze this termwith the intent of driving it toward zero. Clearly, as tl approaches t2, this gravity-loss term approaches zero. In the limitwith tl = t2, the bum is impulsive, and there are no gravity losses. The parameterg, we have no control over, except to note that bums at low altitude (high g) willbe inherently more expensive than bums at high altitude (low g). We can controlthe flight-path angle T during the bum and, if it could be kept to zero, there wouldbe no gravity losses. A finite-duration bum is shown in Fig. 6.11 with dashedlines. Typically, the bum surrounds the impulsive-transfer location (shown withsolid lines), and the finite-bum direction is very close to the impulsive direction.Recall that, for Hohmann transfers, the flight-path angle is zero for both bums. Ifthe impulsive-bum direction is to be maintained during the finite bum, the initialand final values of flight-path angle T cannot be zero. In fact, the initial values ofT are approximately - Y 1 ~ T2 ~ 4~1 ~ 4~2so that the range of T is determined directly by the central angle through whichthe bum occurs. Near the midpoint of the bum arc, T ~ 0 as desired but, at theends, the sin T contribution can be significant. An alternative would be to vary the thrust direction during the bum so that, atany instant of time, T = 0. By definition, the gravity-loss term would be zero in *Note that dm dm dW T = -u -~- = -g0Isp ~ = -Isp ~-
13.
COMPLICATIONS TO IMPULSIVE MANEUVERS 129 ( ii ",. / I . / , •, i/ / //// *Fig. 6.11 Comparison of impulsive- and finite-burn transfers: solid lines = impulsive-burn characteristics; dashed lines = finite-burn characteristics.Eq. (6.5). The fallacy in this approach is that the spacecraft would no longer followthe optimal-thrust direction so that the characteristic AV in Eq. (6.5) would beincreased. One way or another, the penalty must be paid. The factors that lead to high gravity losses are, then, long burn duration (lowthrust/weight), high g (low altitude), and high T (large burn central angle). Thesegravity losses must be added to the impulsive A V to arrive at an effective A V toaccomplish a maneuver. Figure 6.12 shows the total A V required to ascend from alow Earth orbit, LEO (h = 278 km, i = 28.5 deg) to a geosynchronous equatorialorbit, GEO (h = 35786 km, i = 0 deg) as a function of the initial thrust/weightratio. For high thrust/weight ratios, the total A V approaches the impulsive A V(4237 m/s). For low thrust/weight ratios, the gravity loss is obviously significant(over 2200 m/s for T~ W0 = 10-4).Calculating Gravity Losses For the purpose of determining gravity losses, the problem can be roughlydivided into three categories. High thrust (T/Wo.~ (0.5 to 1.0). Here the thrust is the dominant force.Gravity losses can be estimated by using the methods of Robbins 4 or even neglected(impulsive assumption). Low thrust (T/We ~ 10 -2 to 10-1). Here the thrust and gravitational forcesare both important, and assumptions or estimates are difficult. The best method ofsolution is to run an integrated trajectory computer program. Very low thrust (T/W0 ~ 10-5). Here the thrust may be considered as aperturbation to the trajectory. An orbit transfer consists of many orbit revolutions
14.
130 ORBITAL MECHANICS 10 q 5 6 6V (KILONETERS PER SEC) Fig. 6.12 AV requirements to transfer from LEO to GEO.with continuous thrusting. A first-order analytic solution for this case will bedescribed in the next section.Reducing Gravity Losses The obvious method for reducing gravity losses is to install an engine with ahigher thrust. There is, however, another method. A single large (in terms of centralangle traversed) burn can be broken up into a series of smaller burns, separated byone or more revolutions. Consider the transfer of Fig. 6.11. Rather than performa single large burn over the central angle range ¢1 + 4>2, the burn could be splitinto two parts. In the first part, roughly half the total A V1 would be applied, overroughly half the previous central angle range. After a full revolution, the vehiclewould return to very nearly the same perigee location, where the remainder ofthe A V1 would be applied. The net effect is that the central-angle travel (and thecorresponding bounds on T) has been roughly halved. In the gravity-loss terms ofEq. (6.5), this means lower gravity losses. Breaking the burn into more and morepieces, each of which more closely resembles the impulsive case, leads to lowerand lower gravity losses. For the geosynchronous transfer of Fig. 6.12, some cases are plotted in whichthe large perigee burn has been broken into two or four pieces. The A V saving issubstantial. The penalty paid for this is in the form of transfer time. The tradeoffbetween AV savings and transfer time for the case of T/Wo = 0.01 is shownin Fig. 6.13 for the geosynchronous transfer. As the number of perigee burnsincreases, the gravity loss decreases, but the transfer time increases.6.4 Very Low Thrust Transfers If the thrust is small compared to the gravitational force (T/Wo~10-5), then theresulting transfer orbit is a slow spiral outward under continuous thrust. That eachof the many revolutions is nearly circular allows certain simplifying assumptions,
15.
COMPLICATIONS TO IMPULSIVE MANEUVERS 131 a) 1.2 ! V L0.8 0 S S ^0.6 K M / i 8 O . q ................................................................................................... i ............................................... E ! C v ! 0.2 .................................................................................................. i ................................................ ] o i 2 3 q NUMBEROF PERIGEEBURNS ,3 ................................................ i ................................................. i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B I N I T I B L THRUST/HEIGHT = 0 . 0 1 N S 121"................................................................................................... ~ ..................... F / TIB I E B ................................................................................................................... sRH6 7 8 ~ ~ i i 5 2 3 NUMBEROF PERIGEEBURNSFig. 6.13 a) AV loss as a function of number of perigee burns: b) transfer time as afunction of number of perigee burns.which have been used by Edelbaum 5 to develop a first-order analytic solution tothe problem of very low thrust transfers between circular orbits. The given quantities are V0 = initial orbit circular velocity V1 = final orbit circular velocity (x = plane change angle between orbits T = engine thrust
16.
132 ORBITAL MECHANICS Isp = engine specific impulse W1 = payload weight delivered to final orbitThe first step is to compute the A V required to make the transfer. This is given byEdelbaum as A V = Jv0 - 2roY1 cos ~o~ + v~Then, use the rocket equation to find the required weight in the initial orbit (W0) Wo = W1 e x p ( A V/goIsp)where go is the gravitational acceleration at the Earths surface. The propellantrequired for the transfer is simply AW = W0 - W 1The transfer time is, then, ~t = A w/Wwhere ~V = thmst/IspUsing these equations, a first-order estimate of fuel and time requirements for avery low thrust upper-stage vehicle can be obtained. A more complete treatmentof this problem is given in Chapter 14. References 1Edelbaum, T. N., "How Many Impulses?," Aeronautics and Astronautics, Nov. 1967. 2Chu, S. T., Lang, T. J., and Winn, B. E., "A Velocity Matching Technique for ThreeDimensional Orbit Transfer in Conceptual Mission Design," Journal of the AstronauticalSciences, Vol. 26, Oct.-Dec. 1978, pp. 343-368. 3Der, G. J., "Velocity Matching Technique Revisited," AAS/AIAA Astrodynamics Spe-cialist Conference, Lake Tahoe, NV, Aug. 3-5, 1981. 4Robbins, H. M., "Analytical Study of the Impulsive Approximation," AIAA Journal,Vol. 4, Aug. 1966, pp. 1423-1477. 5Edelbaum, T. N., "Propulsion Requirements for Controllable Satellites," ARS Journal,Vol. 31, Aug. 1961, pp. 1079-1089. Problems6.1. For the fixed-impulse transfer example in the text, it is desired to increasethe satellite weight from 907 to 990 kg. If this heavier satellite is placed atop thesame upper-stage vehicle a) Calculate the available AV1 and AV2. b) Using these AV in a Hohmann-type transfer, what plane changes al and ol2must be performed?
17.
COMPLICATIONS TO IMPULSIVE MANEUVERS 133 c) What is the locus of Shuttle orbits (inclination and node) from which thesefixed-impulse Hohmann-type, transfers can be achieved? Sketch in the locus ontoFig. 6.5. d) Are non-Hohmann transfers available? Do they increase the allowable locusof Shuttle parking orbits? e) Briefly describe the launch window available for a Shuttle parking orbit ati = 55 deg.6.2. Instead of using two solid rocket motors to propel the satellite from theShuttle parking orbit to the final orbit, let us now consider a very low thrustion-propulsion engine: Thrust = 4.45 N Isp = 3000 swhere 1 N = 1 kg m/s 2 and the weight of 1 kg is W = mgo = 1 kg x 9.8066m/s 2 = 9.8066 NThe payload weight to be placed on orbit now weighs 1361 kg and consists of thesatellite, the ion engine, and empty propellant tanks. a) For a Shuttle park orbit at i = 55 deg, what is the minimum plane changethat must be performed by the ion upper stage? b) How much A V is required from the ion engine? c) How much gravity loss does this represent? (See example for optimal impul-sive A V.) d) How much propellant must the ion upper stage carry? e) How long will the transfer take?6.3. A payload of space experiments will be delivered to circular low Earth orbit(296-km altitude) by the Space Shuttle. Its final destination is a 741-kin altitudecircular orbit at as high an inclination as possible (preferably above i -= 72 deg).Because of range safety constraints, the highest inclination available for the Shuttleorbit is i = 57 deg. The transfer from the Shuttle orbit to the destination orbit willbe accomplished using two identical solid-propellant motors. These motors arenot restartable, so that one motor will be consumed in leaving the Shuttle orbit andthe second will be consumed in entering the destination orbit. The characteristicsof the motors are as follows: Payload = 1452 kg Solid motor #2 propellant wt -- 1044 kg structure wt -- 136 kg Isp = 285 s C) Solid motor #1 propellant wt = 1044 kg structure wt = 136 kg Isp = 285 s
18.
134 ORBITAL MECHANICSThe required orbital transfer (a Hohmann-type transfer will be used) is summarizedin the following sketch: t~v 2 Shuttle orbit, h = 296 kin, / / ~ ~ 77 8m s = 7 83Hohmann-transferorbit / ~ /hpT R = 296 km, h A T R = 741 km ~ /VeT ~ = 7851 re~s, VATR= 7361 m/sFor this orbital-transfer situation a) What AV(AV1 and AV2) are available from the two solid motors? b) How much plane change (0/1 and 0/2) can be provided by each of the solidmotors when flown on this transfer? c) What is the maximum value of inclination for the destination orbit usingthese motors? Have the experimenters met their goal of i > 72 deg? d) What is the transfer-orbit inclination for this case, and at what latitude mustthe two bums take place? Selected Solutions6.1. a) A V~ = 2048 m/s AV2 = 1779 m/s b)0/1 ~ 0 0/2 = 19.97 deg C) 0/2 ~ 19.97 deg6.2. a) Ai = 0 b) AV = 3865 m/s c) 395 m/s d) mp -----191 kg e) At = 14.7 days6.3. a) zXVI = 894 m/s AV2 = 1412 m/s b) 0/1 = 6.52 deg 0/2 = 10.87 deg c)/max > 72 deg d) itr = 57 + 6.52 = 63.52 deg
Be the first to comment